OCR A Level Chemistry A: Carbon–carbon bond formation (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set gives you practice on section 6.2.4, Carbon–carbon bond formation, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. Between them, the questions reach all four outcomes, 6.2.4(a) to (d), in Module 6 (Organic chemistry and analysis); none of it is labelled AS, since H432 is the full A level. Expect longer, mixed-outcome problems with yield work in the final three questions.

If a reaction is unfamiliar, reread the full 6.2.4 guide or the condensed 6.2.4 notes before you start. Every other H432 section is reached from the H432 hub page, and you can record secure outcomes on the course checklist. For the wider course, use the 10-minute diagnostics.

All data you need for calculations appears in the questions. Use 24.0 dm³ for one mole of any gas at room temperature and pressure.

Questions

1. Cyanide ions react with haloalkanes and with carbonyl compounds.

(a) State what is meant by a nucleophile. [1] (b) Identify the atom in CN⁻ whose lone pair is used to make the new C–C bond. [1]

2. 1-Bromo-2-ethylbutane, (CH₃CH₂)₂CHCH₂Br, is heated under reflux with potassium cyanide in ethanol.

(a) Write a balanced equation using KCN. [1] (b) Name the nitrile, following IUPAC rules. [1] (c) Explain why chemists find this reaction useful when building larger molecules. [1]

3. (Chloromethyl)cyclopentane, C₅H₉CH₂Cl, reacts with sodium cyanide in ethanol.

(a) Name the type of mechanism. [1] (b) Describe the mechanism. For each curly arrow, state its start and end points. Give both products. [3]

4. Cyclohexanecarbaldehyde, C₆H₁₁CHO, reacts with NaCN(aq)/H⁺(aq).

(a) Give the structural formula and the name of the organic product. [2] (b) Describe the mechanism, using H⁺ for the final step. [3]

5. Nonanenitrile, CH₃(CH₂)₇CN, is hydrogenated over a nickel catalyst.

(a) Give a balanced equation. [1] (b) Name the organic product. [1] (c) 5.56 g of nonanenitrile (Mᵣ 139.0) is reduced completely. Calculate the volume of hydrogen used at room temperature and pressure. [2]

6. The product of question 4 is converted into a carboxylic acid.

(a) State the reagent and conditions. [1] (b) Write an equation for the reaction using hydrochloric acid. [2]

7. Benzene reacts with 2-bromo-2-methylbutane, CH₃CH₂C(CH₃)₂Br, in the presence of iron(III) bromide.

(a) Write the overall equation and name the organic product. [2] (b) Write an equation to show how the electrophile forms. [1] (c) Give the equation in which the FeBr₃ catalyst is re-formed. [1]

8. Six reactions are listed.

A: 1-chloropropane with NaOH(aq) B: 2-iodobutane with KCN in ethanol C: propanone with NaBH₄ D: butanal with NaCN(aq)/H⁺(aq) E: benzene with ethanoyl chloride and AlCl₃ F: ethanenitrile with H₂/Ni

Give the letters of the three reactions that form a new carbon–carbon bond. [3]

9. Bromocycloheptane, C₇H₁₃Br, is converted in two steps into cycloheptanecarboxylic acid, C₇H₁₃COOH.

(a) For each step, state the reagent and conditions. Name the intermediate. [3] (b) State which step forms a carbon–carbon bond. [1]

10. Benzene reacts with heptanoyl chloride, CH₃(CH₂)₅COCl, and aluminium chloride.

(a) Write the overall equation for the reaction. [1] (b) Show, with an equation, how AlCl₃ produces the electrophile. [1] (c) 3.90 g of benzene reacts with excess heptanoyl chloride, and 6.97 g of the ketone C₁₃H₁₈O is obtained. Calculate the percentage yield. Mᵣ: benzene 78.0, C₁₃H₁₈O 190.0. [3] (d) Calculate the atom economy for making the ketone. Mᵣ of heptanoyl chloride = 148.5. [2]

11. Compound A, C₇H₁₅Br, has an unbranched carbon chain, and its molecule is symmetrical. A is heated with KCN in ethanol to give B. B is reduced with H₂/Ni to give C. B is heated under reflux with dilute hydrochloric acid to give D.

(a) Give the structural formula and name of A. [1] (b) Give the structural formula and name of B. [2] (c) Give the structural formulae of C and D. [2] (d) 7.16 g of A (Mᵣ 178.9) gives 3.25 g of B (Mᵣ 125.0). Calculate the percentage yield. [3]

12. A chemist plans to make 2-hydroxy-2-phenylbutanoic acid, C₆H₅C(OH)(CH₂CH₃)COOH, from benzene in three steps.

(a) Describe the route. Give the reagents and conditions for each step, the structure of each intermediate, and state which steps form carbon–carbon bonds. [6] (b) The step yields are 72%, 80% and 55%. Calculate the mass of benzene (Mᵣ 78.0) needed to make 2.00 g of the acid (Mᵣ 180.0). [3]

Answers

1. (a) A species able to donate a lone pair of electrons, forming a dative or new covalent bond [1] (b) The carbon atom, which carries a lone pair [1] Examiner insight: “Something attracted to positive charge” is too vague; the idea of donating an electron pair is what earns the mark.

2. (a) (CH₃CH₂)₂CHCH₂Br + KCN → (CH₃CH₂)₂CHCH₂CN + KBr (or with CN⁻ and Br⁻) [1] (b) 3-Ethylpentanenitrile [1] (c) A new C–C bond forms, so the carbon chain is lengthened by one carbon [1] Examiner insight: Number from the nitrile carbon; keeping the haloalkane’s “butane” stem gives the wrong name.

3. (a) Nucleophilic substitution [1] (b) Dipole Cδ+–Clδ− shown; curly arrow starting at the cyanide carbon’s lone pair and ending at the δ+ CH₂ carbon [1]. The C–Cl bonding pair moves onto chlorine (second arrow) [1]. Products: C₅H₉CH₂CN (2-cyclopentylethanenitrile) and Cl⁻ [1] Examiner insight: Arrows must start at a lone pair or a bond; an arrow from the ring or from the minus sign on CN⁻ is unlikely to gain credit.

4. (a) C₆H₁₁CH(OH)CN [1]; 2-cyclohexyl-2-hydroxyethanenitrile [1] (b) First arrow starts at the cyanide carbon’s lone pair and ends at the δ+ C of C=O [1]. Second arrow moves the π pair of C=O onto oxygen, giving the intermediate C₆H₁₁CH(CN)O⁻ [1]. Arrow from an O⁻ lone pair to H⁺, giving the product [1] Examiner insight: Show the negative charge on the oxygen of the intermediate; the protonation arrow then has a clear starting point.

5. (a) CH₃(CH₂)₇CN + 2H₂ → CH₃(CH₂)₇CH₂NH₂ [1] (b) Nonan-1-amine [1] (c) Amount of nitrile = 5.56 ÷ 139.0 = 0.0400 mol, so H₂ = 0.0800 mol [1]; volume = 0.0800 × 24.0 = 1.92 dm³ [1] Examiner insight: Take two moles of H₂ per mole of nitrile from your equation in (a); a 1 : 1 ratio halves the volume.

6. (a) Dilute hydrochloric acid, heated under reflux [1] (b) C₆H₁₁CH(OH)CN + 2H₂O + HCl → C₆H₁₁CH(OH)COOH [1] + NH₄Cl, correctly balanced [1] Examiner insight: The OH group is not changed; only –CN becomes –COOH, and the nitrogen ends up as ammonium chloride.

7. (a) C₆H₆ + CH₃CH₂C(CH₃)₂Br → C₆H₅C(CH₃)₂CH₂CH₃ + HBr [1]; (2-methylbutan-2-yl)benzene (2-methyl-2-phenylbutane) [1] (b) CH₃CH₂C(CH₃)₂Br + FeBr₃ → CH₃CH₂C(CH₃)₂⁺ + FeBr₄⁻ [1] (c) H⁺ + FeBr₄⁻ → FeBr₃ + HBr [1] Examiner insight: The inorganic product here is HBr, not HCl; copy the halogen from the reagents given.

8. B [1], D [1], E [1] Examiner insight: A, C and F change functional groups without changing the carbon skeleton; list only the reactions that join carbon to carbon.

9. (a) Step 1: KCN (or NaCN) in ethanol, heat under reflux [1]. Intermediate: cycloheptanecarbonitrile, C₇H₁₃CN [1]. Step 2: dilute HCl (or H₂SO₄), heat under reflux [1] (b) Step 1 [1] Examiner insight: “Acid” alone is too vague for step 2; name a dilute acid and give reflux.

10. (a) C₆H₆ + CH₃(CH₂)₅COCl → C₆H₅CO(CH₂)₅CH₃ + HCl [1] (b) CH₃(CH₂)₅COCl + AlCl₃ → CH₃(CH₂)₅CO⁺ + AlCl₄⁻ [1] (c) Amount of benzene = 3.90 ÷ 78.0 = 0.0500 mol [1]; theoretical mass = 0.0500 × 190.0 = 9.50 g [1]; yield = 6.97 ÷ 9.50 × 100 = 73.4% [1] (d) Total Mᵣ of reactants = 78.0 + 148.5 = 226.5 [1]; atom economy = 190.0 ÷ 226.5 × 100 = 83.9% [1] Examiner insight: Atom economy uses the balanced equation, not the masses measured; the catalyst is not included because it is regenerated.

11. (a) CH₃CH₂CH₂CHBrCH₂CH₂CH₃, 4-bromoheptane [1] (b) (CH₃CH₂CH₂)₂CHCN [1]; 2-propylpentanenitrile [1] (c) C: (CH₃CH₂CH₂)₂CHCH₂NH₂ [1]; D: (CH₃CH₂CH₂)₂CHCOOH [1] (d) Amount of A = 7.16 ÷ 178.9 = 0.0400 mol [1]; theoretical mass of B = 0.0400 × 125.0 = 5.00 g [1]; yield = 3.25 ÷ 5.00 × 100 = 65.0% [1] Examiner insight: In (b), the longest chain through the nitrile carbon has five carbons, not eight; naming it as an octanenitrile ignores the branch.

12. (a) Step 1: propanoyl chloride, CH₃CH₂COCl, with AlCl₃ (halogen carrier), dry [1]. Intermediate: C₆H₅COCH₂CH₃, 1-phenylpropan-1-one [1]. Step 2: NaCN(aq)/H⁺(aq) [1]. Intermediate: C₆H₅C(OH)(CN)CH₂CH₃, 2-hydroxy-2-phenylbutanenitrile [1]. Step 3: dilute HCl, heat under reflux [1]. Steps 1 and 2 both form C–C bonds [1] (b) Overall yield = 0.72 × 0.80 × 0.55 = 0.3168 [1]; amount of acid = 2.00 ÷ 180.0 = 0.01111 mol [1]; benzene = 0.01111 ÷ 0.3168 × 78.0 = 2.74 g [1] Examiner insight: Plans gain credit step by step, so give reagent, condition and structure for each stage even if one later step is uncertain.

Where marks are usually lost

• Starting the CN⁻ arrow at N rather than at the carbon atom’s lone pair (questions 3 and 4).
• Leaving out the O⁻ intermediate or its protonation in nucleophilic addition.
• Writing “aqueous KCN” for the haloalkane reaction instead of cyanide in ethanol.
• Naming the nitrile from the wrong end or without counting the CN carbon (questions 2 and 11).
• Using one H₂ per nitrile in reduction equations and volume calculations.
• Writing NH₃ instead of NH₄⁺ or NH₄Cl as the hydrolysis by-product.
• Leaving HCl or HBr off a Friedel–Crafts equation, or mixing up the halogen of the carrier.
• Multiplying by the yield instead of dividing when working back to a starting mass (question 12).

Next steps

• Recap reagents and mechanisms: Carbon–carbon bond formation revision notes
• Full explanations and worked examples: Carbon–carbon bond formation study guide
• Yield and atom economy practice: Amount of substance practice
• Check the whole course: all free 10-minute diagnostics
• More OCR chemistry topics: course hub
• Printable checklist: OCR A Level Chemistry checklist
• Book a free trial class: /trial/

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 6.2.4 Carbon–carbon bond formation, learning outcomes (a) to (d). Published by OCR.